Work done by kinetic frictional force

[Jacque77]

Homework Statement

Answer
Chapter 6, Problem 73 Cutnell & Johnson 7th edition
A 356 N force is pulling an 90.2-kg refrigerator across a horizontal surface. The force acts at an angle of 25.6 ° above the surface. The coefficient of kinetic friction is 0.258, and the refrigerator moves a distance of 9.55 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.
(a)
Number
3066.050223272106 Units J
(b)
Number
-1798.985720775539 Units J

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First : sig figs are turned off in my course. The listed answers have to be within 2%. Please no lectures about sig fig. In any other course I obey the rules. ;)

Incorrect.

(a) Number 3066.050223 Units J (is correct)

(b) Number -1964.181362 Units J (is incorrect)

The 3066.050223J work for the part (a) is correct. Part (b) is the incorrect part I need help with.
Wf = Ff*d*cos 25.6 where Ff = μ*m*g
so 0.258*90.2*-9.8*9.55*cos 180 = -2178 J
0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362
I've exceeded the 3 tries and can see the correct answer of -1798.985720775539, but have no idea how that was arrived at. The hint given with the problem says to set ΣFy = 0 before considering the Wf . ??
What am I doing wrong?

Homework Equations

W= F*D cos 25.6
Wf = Ff*d*cos 25.6 where Ff = μ*m*g
FN = mg = weight of the refrigerator

The Attempt at a Solution

W done by pulling force
W= F*D cos 25.6
W= 356N*9.55m*cos 25.6 = 3066.050223J

work done by the frictional force is
.258*90.2*9.8*9.55*cos 180 = -2178J

0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362 is incorrect. The correct answer is -1798.985720775539. ?How?

 

[vector3]

Is there a y componet to the 356 N force as well as an x componet?

 

[Jacque77]

the 90.2 cos 25.6 considers the y. The 9.55 m moved is the x

 

[Doc Al]

Wf = Ff*d*cos 25.6 where Ff = μ*m*g

This is incorrect:
(1) When using W = F*d*cos(theta), theta is the angle between the force and the displacement. In this case the force (friction) is horizontal and so is the displacement.
(2) Ff = μ*N, but N is not simply mg. You need to consider the effect of the vertical component of the applied force.

 

[Jacque77]

The normal reaction force exerted on the refrigerator,
FN = mg - F sin 25.6 = 90.2x9.8 - 356x0.432
= 884 - 154 = 730 N.
Force of friction,
f = μ FN = 0.258 x 730
=188.3 N
along the horizontal surface, but opposite to displacement.

Work done by the frictional force,
Wf = f.S = - f.S cos 180 = -f.S
= - 188.3 x9.55 = - 1798. J

 

[Doc Al]

Now you've got it.